Difficulty finding the sum of a hyperbolic function.

Question

Can someone please point out where I am (If I am) going wrong during the solution process of the following question:

I have been presented with the following :

$$4sinh(2ln(2))-cosh(ln2)$$

and told by my tutor the solution is 10. however I cannot obtain this value, the steps I take are as follows :

$$(4*(e^{ln(2)^2} - e^{-ln(2)^2}/2)) - (e^{ln(2)} + e^{-ln(2)}/2) $$

$$2e^{2ln(2)} - 2e^{-2ln(2)} - 0.5e^{ln(2)} + 0.5e^{-ln(2)} $$

$$8 - 0.5 - 1 + 0.25$$

$$6.75$$

For added clarity I will show a picture of notes I have regarding the question :

I would very much appreciate if someone can help me reach a solution as to why my answer differs from the tutors. Thank you.

Answer 1

The answer should be $6.25$.

\begin{align} & 4 \sinh (2 \ln 2) - \cosh(\ln2 ) \\ =& 2 \left(e^{2\ln2}-e^{-2\ln2} \right) - \frac{e^{\ln2}+e^{-\ln2}}{2}\\ =& 2(4-0.25)-\frac{2+0.5}{2}\\ =& 7.5-1.25 = 6.25. \end{align}

Answer 2

You're correct: $$\begin{align} 4\sinh{2\log{2}} - \cosh{\log{2}} &= \cosh{(\log{2})} \left( 8\sinh{(\log{2})}-1 \right) \\ &= \frac{1}{2}(2+1/2)(4 \cdot 2 - 4 \cdot 1/2-1) \\ &= \frac{5}{4}(8-2-1) \\ &= \frac{25}{4} \end{align}$$